Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
$\int\sqrt{x}\tan\left(2\tan^{-1}\left(\frac{\sqrt{1+x+\sqrt{1}+x}-1}-\sqrt{1+\sqrt{x}-1}}{\sqrt{1+x+\sqrt{1}+x}-1}+\sqrt{1+\sqrt{x}-1}\right)\right)dx$ is equal to $ax^k + K\tan^{-1}\left(\frac{\sqrt{x}}{2}\right) + \frac{a}{\sqrt{1+\sqrt{x}}} + c$ then $a+b$ is equal to (where $a, b, k, a \in \mathbb{R}$)
Step-by-Step Solution
Key Concept: Use the substitution $\sqrt{x} = \tan\theta$ to convert the complicated inverse tangent expression into a manageable trigonometric form.
Substitute $\sqrt{x} = \tan\theta$, so $\frac{1}{2\sqrt{x}}dx = 2\tan\theta\sec^2\theta d\theta$ and $dx = 4\tan\theta\sec^2\theta d\theta$. The integral becomes $\int\tan^2\theta\tan(2\tan^{-1}(\sqrt{1+\tan^2\theta+1}-\sqrt{1+\tan^2\theta-1})) \cdot 4\tan\theta\sec^2\theta d\theta$. Simplify using $\sqrt{1+\tan^2\theta} = \sec\theta$ to get $\int 4\tan^3\theta\sec^2\theta\tan(\frac{\pi}{4}-\frac{\theta}{2})d\theta = \int 4\tan^3\theta\sec^2\theta d\theta$. This evaluates to $\frac{4}{5}\tan^5\theta + c = \frac{4}{5}x^{5/4} + c$.
Correct Answer: I need to find the values of a, b, and k from the given result and then compute a + b.
From the problem, the integral equals:
$$ax^k + K\tan^{-1}\left(\frac{\sqrt{x}}{2}\right) + \frac{a}{\sqrt{1+\sqrt{x}}} +